Lecture Packet #8: Pump Test Analysis

1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet #8: Pump Test Analysis The idea of a pump test is to stress the aquifer by pumping or injecting water and to note the drawdown over space and time. History • • • The earliest model for interpretation of pumping test data was developed by Thiem (1906) (Adolf and Gunther) for o Constant pumping rate o Equilibrium conditions o Confined and unconfined conditions Theis (1935) published the first analysis of transient pump test for o Constant pumping rate o Confined conditions Since then, many methods for analysis of transient well tests have been designed for increasingly complex conditions, including o Aquitard leakage (study of hydrogeology has become more a study of aquitards and less of aquifers) o Aquitard storage o Wellbore storage o Partial well penetration o Anisotropy o Slug tests o Recirculating well tests (water is not removed) It is important to note the assumptions for a given analysis. Steady Radial Flow in a Confined Aquifer Assume: • Aquifer is confined (top and bottom) • Well is pumped at a constant rate • Equilibrium is reached (no drawdown change with time) • Wells are fully screened and is only one pumping 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 1 of 8 Consider Darcy’s law through a cylinder, radius r, with flow toward well. Q=K dh Q dr 2πrb and rearrange as dh = 2πKb r dr Integrate from r1, h1 to r2, h2 h2 r Q 2 dr dh = ∫ 2πKb ∫r1 r h1 h 2 − h1 = T= Q r2 2 π Kb r1 Q r ln 2 2π (h2 − h1 ) r1 or noting that T = Kb this is the Thiem equation Notes • • • on the Thiem equation: Good with any self consistent units L and t If drawdown has stabilized can use any two observation wells Water is not coming from storage (S doesn’t appear) cannot get S from this test • Commonly used in USGS units and log10, T in gpd/ft (gallons per day per foot), Q in gpm (gallons per minute), r and h in ft. T= r 527.7Q log 2 (h2 − h1 ) r1 Specific Capacity of a Well – Roughly estimating T Specific Capacity = Discharge Rate/Drawdown in the well 1. A well is pumped to approximate equilibrium. 2. A good well would be 50 gpm per foot of drawdown, or 20 feet of drawdown at 1,000 gpm. 3. he and re are the head and corresponding distance from a well where drawdown is effectively zero. 4. Specific capacity = T = Q = ln (he − hw ) T 527.7 log re rw 5. Rule of Thumb – T ~ 1,800 x Specfic Capacity 6. What is re? It doesn’t matter that much. log re/rw = 3 re = 1,000 rw log re/rw = 4 re = 10,000 rw 7. Case A Æ T = Specific Capacity [527 x 3] = 1,581 x SC Case B Æ T = Specific Capacity [527 x 4] = 2,108 x SC 8. If you use T ~ 1,800 x Specific Capacity you are not too far off. SC is gpm/ft and T is gpd/ft. 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 2 of 8 Steady Radial Flow in an Unconfined Aquifer Assume: • • • • • Aquifer is unconfined but underlain by an impermeable horizontal unit. Well is pumped at a constant rate Equilibrium is reached (no drawdown change with time) Wells are fully screened and There is only one pumping well Q Observation well Pumping well h1 h2 K r1 r2 aquifer aquiclude Water table before pumping Water table during pumping Radial flow in the unconfined aquifer is given by Q = K (2πrh) dh Q dr and rearrange as hdh = dr 2πK r Integrate from r1, h1 to r2, h2 h2 ∫ r jdh = h1 Q 2 dr 2πK ∫r1 r r h 2 − h1 Q = ln 2 2 2π K r1 2 K= 2 Q π (h2 − h1 ) 2 2 ln r2 r1 or noting that T = Kb this is the Thiem equation for unconfined conditions (K not T, h2 not h, no 2) 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 3 of 8 • • • If the greatest difference in head in the system is < 2%, then you can use the confined equation for an unconfined system. For r < 1.5hmax (the full saturated thickness) there will be errors using this equation because of vertical flow near the well. If difference in head is > 2% but < 25% use the confined equation with the following correction for drawdown (h2 − h1 ) new = ∆h − ∆h 2 “measured” reduced compared to confined aquifer case 2b Transient Pumping Tests ⎡∂ 2h ∂ 2h ⎤ ∂h =T⎢ 2 + 2 ⎥ S ∂t ∂y ⎦ ⎣ ∂x S = Ssb = Storativity [L3/L3] T = Kb = Transmissivity [L2/T] b = aquifer thickness [L] h is head [L] Assume: • • • • • Aquifer is horizontal, confined both top and bottom, infinite in horizontal extent, constant thickness, homogeneous, isotropic. Potentiometric surface is horizontal before pumping, is not changing with time before pumping, all changes due to one pumping well Darcy’s law is valid and groundwater has constant properties Well is fully screened and 100% efficient Constant pumping from a well in such a situation is radial, and horizontal, where only 1 space dimension is needed S ∂h ∂ 2 h 1 ∂h = + T ∂t ∂t 2 r ∂r r = x2 + y2 r Q Observation well Pumping well Cone of Depression Drawdown aquiclude b r h(r,t) T and S aquifer aquiclude 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 4 of 8 We want a solution that gives h(r,t) after we start pumping. To solve equation we need initial conditions (ICs) and boundary conditions (BCs) IC: h(r,0) = h0 Q ⎛ ∂h ⎞ for t > 0 ⎟= ⎝ ∂r ⎠ 2πT BCs: h( ∞ ,t)=h0 and limrÆ0 ⎜ r which is just the application of Darcy’s law at the well Theis Equation – transient radial flow 1935 – C.V. Theis solves this equation (with C.I. Lubin from heat conduction) ∞ h0 − h(r , t ) = Q e − u du 4πT ∫u u exponential integral in math tables but for our case it is “well function” Å where u= r 2S 4Tt h0 − h(r , t ) = Q W (u) , well function is W(u) 4πT Determining T and S from Pumping Test Inverse method: Use solution to the PDE to identify the parameter values by matching simulated and observed heads (dependent variables); e.g., measure aquifer drawdown response given a known pumping rate and get T and S. 1. Identify pumping well and observation wells and their conditions (e.g., fully screened). 2. Determine aquifer type and make a quick estimate to predict what you think will happen during pumping test. 3. Theis Method: Arrange Theis equation as follows: ⎡114.6Q ⎤ ∆h = ⎢ W (u) (in USGS units) and ⎣ T ⎥⎦ ⎡ 1 . 87 Sr r2 ⎡ T ⎤ t = ⎢ u =⎢ Æ ⎥ T t ⎣1.87 S ⎦ ⎣ ⎡114.6Q ⎤ log ∆h = log ⎢ + logW (u) ⎣ T ⎥⎦ 2 ⎤1 ⎥ ⎦u ⎡1.87 Sr 2 ⎤ 1 log t = log ⎢ ⎥ + log u ⎣ T ⎦ 4. Plot the well function W(u) versus 1/u on log-log paper. (this is called a type curve) 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 5 of 8 5. Plot drawdown vs. time on log-log paper of same scale. (this is from data at a single observation well) 6. Superimpose the field curve on the type curve, keeping the axes parallel. Adjust the curves so that most of the data fall on the type curve. You trying to get the constants (bracketed terms) that make the type curve axes translate into your axes. 7. Select a match point (any convenient point will do like W(u) = 1.0 and 1/u = 1.0), and read off the values for W(u) and 1/u. Then read off the values for drawdown and t. 8. Compute the values of T and S from: Using USGS Units Using Self-Consistent Units 114.6QW (u ) (h0 − h) Ttu S= 1.87r 2 T= T= QW (u ) 4π (h0 − h) 4Ttu S= 2 r If in USGS units of drawdown (ft), Q (gpm), T (gpd/ft), r (ft), t (days), S (decimal fraction). h0 − h(r , t ) = u= Q W (u) 4πT 1.87r 2 S Tt To predict drawdown – drawdown vs. distance or time • Put in r, S, T, t and solve for u • Find W(u) based on u and table • Multiply Q/T and factor The analytic solution describes: • Geometric characteristics to the cone of depression, steepening toward the well • For a given aquifer cone of depression increases in depth and extent with time • Drawdown at a time and location increases linearly with pumping rate • Drawdown at a time and location is greater for smaller T • Drawdown at a time and location is greater for lower S 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 6 of 8 Modified Nonequilibrium Solution – Jacob Method Recall from Theis solution: ∞ h0 − h(r , t ) = Q e − u du 4πT ∫u u C.E. Jacob noted that the well function can be represented by a series. h0 − h(r , t ) = Q 4πT ⎤ ⎡ u2 u3 u4 u u − 0 . 5772 − ln + − + − + ....⎥ ⎢ 2 ⋅ 2! 3 ⋅ 3! 4 ⋅ 4! ⎦ ⎣ For small values of r and large values of t, u becomes small. (valid when u < 0.01). Then most terms can be dropped leaving: Q ⎡ r 2S ⎤ h0 − h(r , t ) = ⎥ ⎢− 0.5772 − ln 4πT ⎣ 4Tt ⎦ Noting that –ln u = ln 1/u and ln 1.78 = 0.5772 h0 − h(r , t ) = Q 4πT ⎡ 2.25Tt ⎤ ⎢⎣ln r 2 S ⎥⎦ And ln u = 2.3 log u h0 − h(r , t ) = 2.30Q ⎡ 2.25Tt ⎤ log10 2 ⎥ ⎢ 4πT ⎣ r S ⎦ Since Q, r, T, and S are constants, drawdown vs. log t should plot as a straight line. In USGS units: h0 − h(r , t ) = 264Q ⎡ 0.3Tt ⎤ log10 2 ⎥ ⎢ 4πT ⎣ r S ⎦ Over one log cycle you get change in drawdown From t1 to t2 : ∆[h0 − h] = T= 264Q [log10 10] ∆[h0 − h] t2 ⎤ 264Q ⎡ ⎢log10 ⎥ T ⎣ t1 ⎦ (USGS units) 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 7 of 8 T= 2.3Q 4π∆[h0 − h] (self-consistent units) Procedure 1. Plot on semi-log paper – t on log scale and drawdown on arithmetic scale. 2. Pick off two values of time and the corresponding values of drawdowns (over one log cycle make it easy). 3. Solve for T (just a function of Q and ∆h) 4. Consider basic solution when drawdown is zero. 0 = h0 − h = 2.25Tt 0 ⎤ 2.30Q ⎡ log10 ⎢ 4πT ⎣ r 2 S ⎥⎦ 2.25Tt 0 ⎤ 2.25Tt 0 ⎡ 0 = ⎢log10 Æ = 1 or r 2S r 2 S ⎥⎦ ⎣ S= 2.25Tt 0 r2 where t0 is the time intercept at which drawdown is zero in USGS units. S= 2.25Tt 0 r2 where t0 is the time intercept in days and the straight line intersects the zerodrawdown axis 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 8 Page 8 of 8